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NCERT Solutions · Class 11 Maths · Chapter 12

Chapter 12: Limits and Derivatives (Calculus)

Step-by-step solutions to all exercises of Chapter 12, Limits and Derivatives (NCERT Class 11 Maths, 2026-27): evaluating limits, standard limits, derivatives from first principles and by rules, and the Miscellaneous Exercise. All 72 questions are answered, with the key answer highlighted.

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$lim_{x \to a} \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}$ , $lim_{x \to 0} \frac{s i n x}{x} = 1$ , $lim_{x \to 0} \frac{1 - c o s x}{x} = 0$ . First principle: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ . Product rule $(uv)^{'} = u^{'} v + u v^{'}$ ; quotient rule $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ .

Exercise 12.1

Evaluate: lim_(x→3)(x + 3)

Solution

No 0/0, so substitute. 3 + 3 = 6.

Evaluate: lim_(x→π)(x − 22/7)

Solution

No 0/0, so substitute. π − 22/7.

π − 22/7

Evaluate: lim_(r→1) πr²

Solution

No 0/0, so substitute. π.

Evaluate: lim_(x→4)(4x + 3)/(x − 2)

Solution

No 0/0, so substitute. 19/2 = 19/2.

19/2

Evaluate: lim_(x→−1)(x¹⁰ + x⁵ + 1)/(x − 1).

Solution

No 0/0, so substitute. (1 − 1 + 1)/(−2) = −½.

−½

Evaluate: lim_(x→0) [(x + 1)⁵ − 1]/x.

Solution

Put y = x + 1 (y → 1): lim (y⁵ − 1⁵)/(y − 1) = 5(1)⁴ = 5.

Evaluate: lim_(x→2)(3x² − x − 10)/(x² − 4).

Solution

0/0. (3x + 5)(x − 2)/((x − 2)(x + 2)) = (3x + 5)/(x + 2) → 11/4.

11/4

Evaluate: lim_(x→3)(x⁴ − 81)/(2x² − 5x − 3).

Solution

(x − 3)(x + 3)(x² + 9)/((x − 3)(2x + 1)) → 6 × 18/7 = 108/7.

108/7

Evaluate: lim_(x→0)(ax + b)/(cx + 1).

Solution

Substitute: b.

Evaluate: lim_(z→1)(z^(1/3) − 1)/(z^(1/6) − 1).

Solution

Divide top and bottom by (z − 1): [(z^(1/3) − 1)/(z − 1)] ÷ [(z^(1/6) − 1)/(z − 1)] → (1/3) ÷ (1/6) = 2.

Evaluate: lim_(x→1)(ax² + bx + c)/(cx² + bx + a), a + b + c ≠ 0.

Solution

(a + b + c)/(c + b + a) = 1.

Evaluate: lim_(x→−2)(1/x + 1/2)/(x + 2).

Solution

1/x + 1/2 = (x + 2)/(2x); divide by (x + 2): 1/(2x) → −¼.

−¼

Evaluate: lim_(x→0) sin ax/bx

Solution

(a/b) · sin ax/(ax) → a/b.

a/b

Evaluate: lim_(x→0) sin ax/sin bx

Solution

[sin ax/(ax)] · [bx/sin bx] · (a/b) → a/b.

a/b

Evaluate: lim_(x→π) sin(π − x)/[π(π − x)]

Solution

put y = π − x → 0: (1/π) · sin y/y → 1/π.

1/π

Evaluate: lim_(x→0) cos x/(π − x).

Solution

substitute: 1/π.

1/π

Evaluate: lim_(x→0)(cos 2x − 1)/(cos x − 1).

Solution

cos 2x − 1 = −2 sin²x, cos x − 1 = −2 sin²(x/2): ratio = sin²x/sin²(x/2) = [sin x/x]² · [(x/2)/sin(x/2)]² · 4 → 4.

Evaluate: lim_(x→0)(ax + x cos x)/(b sin x).

Solution

x(a + cos x)/(b sin x) = (1/b) · (x/sin x) · (a + cos x) → (a + 1)/b.

(a + 1)/b

Evaluate: lim_(x→0) x sec x.

Solution

x/cos x → 0/1 = 0.

Evaluate: lim_(x→0)(sin ax + bx)/(ax + sin bx), a, b, a + b ≠ 0.

Solution

Divide by x: [a · sin ax/(ax) + b]/[a + b · sin bx/(bx)] → (a + b)/(a + b) = 1.

Evaluate: lim_(x→0)(cosec x − cot x).

Solution

(1 − cos x)/sin x = 2 sin²(x/2)/(2 sin(x/2) cos(x/2)) = tan(x/2) → 0.

Evaluate: lim_(x→π/2) tan 2x/(x − π/2).

Solution

Put y = x − π/2 → 0: tan(2y + π) = tan 2y; tan 2y/y = 2 · tan 2y/(2y) → 2.

f(x) = 2x + 3 (x ≤ 0), 3(x + 1) (x > 0). Find lim_(x→0) f and lim_(x→1) f.

Solution

At 0: LHL = 3, RHL = 3(1) = 3 ⇒ lim = 3. At 1 (only the second piece): 3(2) = 6.

lim = 3; 6

f(x) = x² − 1 (x ≤ 1), −x² − 1 (x > 1). Find lim_(x→1) f.

Solution

LHL = 0, RHL = −2: unequal, so the limit does not exist.

the limit does not exist

25–26

Evaluate: lim_(x→0) f for (25) f = |x|/x (x ≠ 0), 0 (x = 0) (26) f = x/|x| (x ≠ 0), 0 (x = 0).

Solution

For x < 0 the value is −1, for x > 0 it is 1: LHL = −1 ≠ RHL = 1 in both cases: no limit.

no limit

Evaluate: lim_(x→5) f, f(x) = |x| − 5.

Solution

Near 5, |x| = x: lim (x − 5) = 0.

f = a + bx (x < 1), 4 (x = 1), b − ax (x > 1); lim_(x→1) f = f(1). Find a, b.

Solution

LHL = a + b = 4 and RHL = b − a = 4 ⇒ a = 0, b = 4.

a = 0, b = 4

f(x) = (x − a₁)(x − a₂)…(x − aₙ). Find lim_(x→a1) f and lim_(x→a) f (a ≠ a_(i)).

Solution

A polynomial: substitute. lim_(x→a1) f = 0; lim_(x→a) f = (a − a₁)(a − a₂)…(a − aₙ).

0; (a − a₁)(a − a₂)…(a − aₙ)

f = |x| + 1 (x < 0), 0 (x = 0), |x| − 1 (x > 0). For which a does lim_(x→a) f exist?

Solution

At 0: LHL = 1, RHL = −1: no limit. For a < 0, f = −x + 1 near a; for a > 0, f = x − 1: polynomials, limits exist. So the limit exists for every a ≠ 0.

the limit exists for every a ≠ 0

lim_(x→1)(f(x) − 2)/(x² − 1) = π. Find lim_(x→1) f(x).

Solution

f(x) − 2 = [(f(x) − 2)/(x² − 1)] · (x² − 1) → π × 0 = 0 ⇒ lim f(x) = 2.

lim f(x) = 2

f = mx² + n (x < 0), nx + m (0 ≤ x ≤ 1), nx³ + m (x > 1). For which integers m, n do lim_(x→0) f and lim_(x→1) f both exist?

Solution

At 0: LHL = n, RHL = m ⇒ need m = n. At 1: LHL = n + m, RHL = n + m: always exists. So both exist when m = n (any integers).

m = n

Exercise 12.2

Derivative of: x² − 2 at x = 10

Solution

lim [(10 + h)² − 100]/h = lim (20 + h) = 20.

Derivative of: x at x = 1

Solution

lim [(1 + h) − 1]/h = 1.

Derivative of: 99x at x = 100.

Solution

lim [99(100 + h) − 9900]/h = 99.

From the first principle: (i) x³ − 27 (ii) (x − 1)(x − 2) (iii) 1/x² (iv) (x + 1)/(x − 1).

Solution

(i) [(x + h)³ − x³]/h = 3x² + 3xh + h² → 3x². (ii) f = x² − 3x + 2: [2xh + h² − 3h]/h → 2x − 3. (iii) [1/(x + h)² − 1/x²]/h = −(2x + h)/(x²(x + h)²) → −2/x³. (iv) [(x + h + 1)/(x + h − 1) − (x + 1)/(x − 1)]/h = −2/((x + h − 1)(x − 1)) → −2/(x − 1)².

(i) 3x²; (ii) 2x − 3; (iii) −2/x³; (iv) −2/(x − 1)²

f(x) = x¹⁰⁰/100 + x⁹⁹/99 + … + x²/2 + x + 1. Prove f′(1) = 100 f′(0).

Solution

f′(x) = x⁹⁹ + x⁹⁸ + … + x + 1. f′(1) = 100 (one hundred 1s), f′(0) = 1. Hence f′(1) = 100 f′(0).

f′(1) = 100 f′(0)

Derivative of xⁿ + axⁿ⁻¹ + a²xⁿ⁻² + … + aⁿ⁻¹x + aⁿ.

Solution

nxⁿ⁻¹ + a(n − 1)xⁿ⁻² + a²(n − 2)xⁿ⁻³ + … + aⁿ⁻¹.

nxⁿ⁻¹ + a(n − 1)xⁿ⁻² + a²(n − 2)xⁿ⁻³ + … + aⁿ⁻¹

Find the derivative of: (i) (x − a)(x − b) (ii) (ax² + b)² (iii) (x − a)/(x − b).

Solution

(i) product rule: (x − b) + (x − a) = 2x − a − b. (ii) a²x⁴ + 2abx² + b² ⇒ 4a²x³ + 4abx. (iii) [(x − b) − (x − a)]/(x − b)² = (a − b)/(x − b)².

(i) 2x − a − b; (ii) 4a²x³ + 4abx; (iii) (a − b)/(x − b)²

Derivative of (xⁿ − aⁿ)/(x − a).

Solution

Quotient rule: [nxⁿ⁻¹(x − a) − (xⁿ − aⁿ)]/(x − a)² = (nxⁿ − anxⁿ⁻¹ − xⁿ + aⁿ)/(x − a)².

(nxⁿ − anxⁿ⁻¹ − xⁿ + aⁿ)/(x − a)²

Find the derivative of: (i) 2x − 3/4 (ii) (5x³ + 3x − 1)(x − 1) (iii) x⁻³(5 + 3x) (iv) x⁵(3 − 6x⁻⁹) (v) x⁻⁴(3 − 4x⁻⁵) (vi) 2/(x + 1) − x²/(3x − 1).

Solution

(i) 2. (ii) (15x² + 3)(x − 1) + (5x³ + 3x − 1) = 20x³ − 15x² + 6x − 4. (iii) 5x⁻³ + 3x⁻² ⇒ −15x⁻⁴ − 6x⁻³ = −3(5 + 2x)/x⁴. (iv) 3x⁵ − 6x⁻⁴ ⇒ 15x⁴ + 24/x⁵. (v) 3x⁻⁴ − 4x⁻⁹ ⇒ −12/x⁵ + 36/x¹⁰. (vi) −2/(x + 1)² − [2x(3x − 1) − 3x²]/(3x − 1)² = −2/(x + 1)² − x(3x − 2)/(3x − 1)².

(i) 2; (ii) 20x³ − 15x² + 6x − 4; (iii) −3(5 + 2x)/x⁴; (iv) 15x⁴ + 24/x⁵; (v) −12/x⁵ + 36/x¹⁰; (vi) −2/(x + 1)² − x(3x − 2)/(3x − 1)²

Derivative of cos x from the first principle.

Solution

[cos(x + h) − cos x]/h = −2 sin(x + h/2) sin(h/2)/h = −sin(x + h/2) · [sin(h/2)/(h/2)] → −sin x.

−sin x

Find the derivative of: (i) sin x cos x (ii) sec x (iii) 5 sec x + 4 cos x (iv) cosec x (v) 3 cot x + 5 cosec x (vi) 5 sin x − 6 cos x + 7 (vii) 2 tan x − 7 sec x.

Solution

(i) cos²x − sin²x = cos 2x. (ii) (1/cos x)′ = sin x/cos²x = sec x tan x. (iii) 5 sec x tan x − 4 sin x. (iv) (1/sin x)′ = −cos x/sin²x = −cosec x cot x. (v) −3 cosec²x − 5 cosec x cot x. (vi) 5 cos x + 6 sin x. (vii) 2 sec²x − 7 sec x tan x.

(i) cos 2x; (ii) sec x tan x; (iii) 5 sec x tan x − 4 sin x; (iv) −cosec x cot x; (v) −3 cosec²x − 5 cosec x cot x; (vi) 5 cos x + 6 sin x; (vii) 2 sec²x − 7 sec x tan x

Miscellaneous Exercise on Chapter 12

From the first principle: (i) −x (ii) (−x)⁻¹ (iii) sin(x + 1) (iv) cos(x − π/8).

Solution

(i) [−(x + h) + x]/h = −1. (ii) f = −1/x: [−1/(x + h) + 1/x]/h = 1/(x(x + h)) → 1/x². (iii) [sin(x + h + 1) − sin(x + 1)]/h = 2 cos(x + 1 + h/2) sin(h/2)/h → cos(x + 1). (iv) similarly → −sin(x − π/8).

(i) −1; (ii) 1/x²; (iii) cos(x + 1); (iv) −sin(x − π/8)

Find the derivative of: x + a

Solution

Find the derivative of: (px + q)(r/x + s)

Solution

= pr + psx + qr/x + qs ⇒ ps − qr/x².

ps − qr/x²

Find the derivative of: (ax + b)(cx + d)²

Solution

a(cx + d)² + 2c(ax + b)(cx + d) = (cx + d)(3acx + ad + 2bc).

(cx + d)(3acx + ad + 2bc)

Find the derivative of: (ax + b)/(cx + d)

Solution

[a(cx + d) − c(ax + b)]/(cx + d)² = (ad − bc)/(cx + d)².

(ad − bc)/(cx + d)²

Find the derivative of: (1 + 1/x)/(1 − 1/x)

Solution

= (x + 1)/(x − 1) ⇒ −2/(x − 1)², x ≠ 0, 1.

−2/(x − 1)²

Find the derivative of: 1/(ax² + bx + c).

Solution

−(2ax + b)/(ax² + bx + c)².

−(2ax + b)/(ax² + bx + c)²

Find the derivative of: (ax + b)/(px² + qx + r)

Solution

[a(px² + qx + r) − (ax + b)(2px + q)]/(px² + qx + r)² = (−apx² − 2bpx + ar − bq)/(px² + qx + r)².

(−apx² − 2bpx + ar − bq)/(px² + qx + r)²

Find the derivative of: (px² + qx + r)/(ax + b).

Solution

[(2px + q)(ax + b) − a(px² + qx + r)]/(ax + b)² = (apx² + 2bpx + bq − ar)/(ax + b)².

(apx² + 2bpx + bq − ar)/(ax + b)²

Find the derivative of: a/x⁴ − b/x² + cos x

Solution

−4a/x⁵ + 2b/x³ − sin x.

−4a/x⁵ + 2b/x³ − sin x

Find the derivative of: 4√x − 2

Solution

4 · ½x^(−1/2) = 2/√x.

2/√x

Find the derivative of: (ax + b)ⁿ

Solution

na(ax + b)ⁿ⁻¹.

na(ax + b)ⁿ⁻¹

Find the derivative of: (ax + b)ⁿ(cx + d)ᵐ

Solution

(ax + b)ⁿ⁻¹(cx + d)ᵐ⁻¹[na(cx + d) + mc(ax + b)].

(ax + b)ⁿ⁻¹(cx + d)ᵐ⁻¹[na(cx + d) + mc(ax + b)]

Find the derivative of: sin(x + a).

Solution

sin x cos a + cos x sin a ⇒ cos(x + a).

cos(x + a)

Find the derivative of: cosec x cot x

Solution

(−cosec x cot x) cot x + cosec x(−cosec²x) = −cosec³x − cosec x cot²x.

−cosec³x − cosec x cot²x

Find the derivative of: cos x/(1 + sin x)

Solution

[−sin x(1 + sin x) − cos²x]/(1 + sin x)² = −(1 + sin x)/(1 + sin x)² = −1/(1 + sin x).

−1/(1 + sin x)

Find the derivative of: (sin x + cos x)/(sin x − cos x)

Solution

[−(sin x − cos x)² − (sin x + cos x)²]/(sin x − cos x)² = −2/(sin x − cos x)².

−2/(sin x − cos x)²

Find the derivative of: (sec x − 1)/(sec x + 1).

Solution

[sec x tan x(sec x + 1) − (sec x − 1) sec x tan x]/(sec x + 1)² = 2 sec x tan x/(sec x + 1)².

2 sec x tan x/(sec x + 1)²

Find the derivative of: sinⁿx

Solution

By induction with the product rule (sinⁿx = sinⁿ⁻¹x · sin x): n sinⁿ⁻¹x cos x.

n sinⁿ⁻¹x cos x

Find the derivative of: (a + b sin x)/(c + d cos x)

Solution

[b cos x(c + d cos x) + d sin x(a + b sin x)]/(c + d cos x)² = (bc cos x + ad sin x + bd)/(c + d cos x)².

(bc cos x + ad sin x + bd)/(c + d cos x)²

Find the derivative of: sin(x + a)/cos x

Solution

[cos(x + a) cos x + sin(x + a) sin x]/cos²x = cos a/cos²x.

cos a/cos²x

Find the derivative of: x⁴(5 sin x − 3 cos x).

Solution

4x³(5 sin x − 3 cos x) + x⁴(5 cos x + 3 sin x).

4x³(5 sin x − 3 cos x) + x⁴(5 cos x + 3 sin x)

Find the derivative of: (x² + 1) cos x

Solution

2x cos x − (x² + 1) sin x.

2x cos x − (x² + 1) sin x

Find the derivative of: (ax² + sin x)(p + q cos x)

Solution

(2ax + cos x)(p + q cos x) − q sin x(ax² + sin x).

(2ax + cos x)(p + q cos x) − q sin x(ax² + sin x)

Find the derivative of: (x + cos x)(x − tan x)

Solution

(1 − sin x)(x − tan x) − (x + cos x) tan²x (as 1 − sec²x = −tan²x).

(1 − sin x)(x − tan x) − (x + cos x) tan²x

Find the derivative of: (4x + 5 sin x)/(3x + 7 cos x).

Solution

[(4 + 5 cos x)(3x + 7 cos x) − (4x + 5 sin x)(3 − 7 sin x)]/(3x + 7 cos x)² = (35 + 28 cos x − 15 sin x + 15x cos x + 28x sin x)/(3x + 7 cos x)².

(35 + 28 cos x − 15 sin x + 15x cos x + 28x sin x)/(3x + 7 cos x)²

Find the derivative of: x² cos(π/4)/sin x

Solution

cos(π/4) is a constant: cos(π/4) · x(2 sin x − x cos x)/sin²x.

cos(π/4) · x(2 sin x − x cos x)/sin²x

Find the derivative of: x/(1 + tan x)

Solution

[(1 + tan x) − x sec²x]/(1 + tan x)².

[(1 + tan x) − x sec²x]/(1 + tan x)²

Find the derivative of: (x + sec x)(x − tan x)

Solution

(1 + sec x tan x)(x − tan x) + (x + sec x)(1 − sec²x).

(1 + sec x tan x)(x − tan x) + (x + sec x)(1 − sec²x)

Find the derivative of: x/sinⁿx.

Solution

[sinⁿx − nx sinⁿ⁻¹x cos x]/sin²ⁿx = (sin x − nx cos x)/sinⁿ⁺¹x.

(sin x − nx cos x)/sinⁿ⁺¹x

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